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cutiecomittee123

  • one year ago

How do i solve this system of equations? y=(x+1)^2-2 and y-2x-3

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  1. whpalmer4
    • one year ago
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    is the second equation \[y=2x-3\]?

  2. whpalmer4
    • one year ago
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    Well, let's assume that it is. \[y = (x+1)^2-2\]\[y=2x-3\] We can solve this by substitution. Everywhere you see \(y\) in the first equation, replace it with the right hand side of the second equation (which is also equal to \(y\)). This gives us an equation all in terms of just the variable \(x\): \[2x-3 = (x+1)^2-2\]Now you can expand that out: \[2x-3 = (x+1)(x+1)-2\]\[2x-3=x^2+1x+1x+1*1-2\]\[2x-3=x^2+2x-1\]and get all the terms on one side and you have a quadratic to solve. You know how to do that?

  3. cutiecomittee123
    • one year ago
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    The second equation is y-2x=3

  4. whpalmer4
    • one year ago
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    Okay, can you solve \[y-2x=3\] for \(y\)?

  5. cutiecomittee123
    • one year ago
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    y=3+2x

  6. whpalmer4
    • one year ago
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    Right. So now we do the same thing I did, except with the correct equation. \[y = (x+1)^2+2\]\[y = 2x+3\] \[2x+3 = (x+1)^2 + 2\] Expand both sides, then combine all the terms on one side and solve the quadratic.

  7. cutiecomittee123
    • one year ago
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    wait it should be 3+2x=(x+1)^2-2

  8. whpalmer4
    • one year ago
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    doesn't \[2x+3 = 3+2x\]?

  9. whpalmer4
    • one year ago
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    sorry, I switched a sign, that's what you were pointing out, my bad!

  10. whpalmer4
    • one year ago
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    \[3+2x = (x+1)^2-2\]

  11. cutiecomittee123
    • one year ago
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    yes but the original equation is y=(x+1)^2-2 and if we are plugging in y , and y=2x+3

  12. cutiecomittee123
    • one year ago
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    Yeah thats what I got :))

  13. whpalmer4
    • one year ago
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    Glad one of us is paying close attention :-)

  14. cutiecomittee123
    • one year ago
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    so now i get 2x+3=x^2+1+2

  15. cutiecomittee123
    • one year ago
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    thats after expanding the equation

  16. whpalmer4
    • one year ago
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    close... \[(x+1)^2 = (x+1)(x+1) = x*x + 1*x + 1*x + 1*1 = x^2 + 2x + 1\] so [2x+3 = x^2+2x+1 - 2\]\[2x+3 = x^2+2x-1\] right?

  17. cutiecomittee123
    • one year ago
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    yes got that

  18. whpalmer4
    • one year ago
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    okay, get all the terms collected on one side, what do you get?

  19. cutiecomittee123
    • one year ago
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    0=x^2-4

  20. whpalmer4
    • one year ago
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    Okay, and what are the solutions to that equation?

  21. whpalmer4
    • one year ago
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    That's a simple one to solve: \[x^2-4 = 0\]\[x^2 = 4\]Take the square root of both sides and you see \[x=\pm 2\] Have to be sure to include the negative square root or we won't get all of the answers! Now we need to find the values of \(y\) that go along with those values of \(x\). We can use either equation, so might as well use the simpler one: \[y=2x+3\] Plug in both values of \(x\) and get the corresponding values of \(y\).

  22. cutiecomittee123
    • one year ago
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    okay just a sec

  23. cutiecomittee123
    • one year ago
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    I get (2,7) and (-2,-1)

  24. whpalmer4
    • one year ago
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    So do I!

  25. whpalmer4
    • one year ago
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    and we can easily test them: \[2(2)+3 = (2+1)^2-2\]\[7 = 9-2\checkmark\] \[2(-2)+3 = (-2+1)^2-2\]\[-1=1-2\checkmark\]

  26. whpalmer4
    • one year ago
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    So we can do this any time we can rearrange the stuff in one equation to give us something we can substitute into the other one.

  27. cutiecomittee123
    • one year ago
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    sweet thanks

  28. whpalmer4
    • one year ago
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    Here's a tip that doesn't really matter here so much, but does in problems that you will encounter: always try your answers out in the original equation and make sure they work! Whenever you have to square both sides of the equation, you can get what called "spurious" solutions which come out just the same way as the real ones do, but don't actually make the equations true when you plug the numbers in. Rather than remembering exactly what kinds of problems require you to do this, I prefer to simply always check my answers!

  29. anonymous
    • one year ago
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    Cool! @whpalmer4

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